A toothed wheel of diameter $50{\text{ cm}}$ is attached to a smaller wheel of diameter $30{\text{ cm}}$. How many revolutions will the smaller wheel make when the larger one makes $15$ revolutions?
A. $25{\text{ revolutions}}$
B. $20{\text{ revolutions}}$
C. ${\text{1 revolution}}$
D. $10{\text{ revolutions}}$
Answer
570.9k+ views
Hint The distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$ .
The distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different.
Complete step by step answer
As given in the question that the toothed wheel of diameter $50{\text{ cm}}$ is attached to a smaller wheel of diameter $30{\text{ cm}}$ .
Let the number of revolutions smaller wheel will make when the larger one makes $15$ revolutions be $n$
We know that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$
So, the distance covered by larger and smaller wheels in one revolution will be $50\pi $ and $30\pi $ respectively.
Now, the distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different. So,
${\text{Distance covered by larger wheel in 15 revolutions }} = {\text{ Distance covered by smaller wheel in n revolutions}}$After substituting the values we have
$15 \times 50\pi = n \times 30\pi $
On simplifying we have
$n = \dfrac{{15 \times 50}}{{30}} = 25$
Therefore, the smaller wheel will make 25 revolutions when the larger one makes 15 revolutions.
Hence, option A is correct.
Note We have used the fact that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$ . Although it is applicable only when the wheel is in uniform circular motion and there is no angular acceleration. In uniform circular motion, the magnitude of velocity of each particle in motion is constant throughout the motion.
Angular acceleration is basically the time rate of change of angular velocity and the angular velocity is the rate of change of angular displacement.
The distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different.
Complete step by step answer
As given in the question that the toothed wheel of diameter $50{\text{ cm}}$ is attached to a smaller wheel of diameter $30{\text{ cm}}$ .
Let the number of revolutions smaller wheel will make when the larger one makes $15$ revolutions be $n$
We know that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$
So, the distance covered by larger and smaller wheels in one revolution will be $50\pi $ and $30\pi $ respectively.
Now, the distance covered by large and small wheels will remain the same although their number of revolutions to cover the distance will be different. So,
${\text{Distance covered by larger wheel in 15 revolutions }} = {\text{ Distance covered by smaller wheel in n revolutions}}$After substituting the values we have
$15 \times 50\pi = n \times 30\pi $
On simplifying we have
$n = \dfrac{{15 \times 50}}{{30}} = 25$
Therefore, the smaller wheel will make 25 revolutions when the larger one makes 15 revolutions.
Hence, option A is correct.
Note We have used the fact that the distance covered by a wheel of diameter $d$ is equal to its circumference i.e. $\pi d$ . Although it is applicable only when the wheel is in uniform circular motion and there is no angular acceleration. In uniform circular motion, the magnitude of velocity of each particle in motion is constant throughout the motion.
Angular acceleration is basically the time rate of change of angular velocity and the angular velocity is the rate of change of angular displacement.
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